Key takeaway: An inverse function is a function that reverses another function, taking each output back to its original input, so that applying both in turn returns the starting number.
Grade 9 – Algebra 1F-BF.B.4aFunctions and Graphs
Definition of Inverse Function
An inverse function undoes what the original function does. If \(f\) turns an input \(x\) into an output \(y\), then its inverse turns \(y\) back into \(x\).
Think of a machine that doubles a number and adds 7. The inverse machine subtracts 7, then halves. For a straight-line rule, the recipe is to trade the places of \(x\) and \(y\), then isolate \(y\) again. Not every function can be reversed: each output must come from exactly one input.
Formula and notation
The inverse of \(f\) is written \(f^{-1}\). The \(-1\) is not an exponent. The defining property is:
\[ f^{-1}\!(f(x)) = x \quad \text{and} \quad f\!(f^{-1}(x)) = x \]
On a graph, the points \((a, b)\) on \(f\) become \((b, a)\) on \(f^{-1}\).
Key facts
- The graph of an inverse function is the mirror image of the original across y = x.
- Domain and range swap: the outputs of f are the inputs of its inverse.
- The small -1 in f^{-1} labels the inverse; it never signals a reciprocal.
- Only a one-to-one function can be reversed, since every output needs a single source.
Inverse Function example
Cab fares in a small town follow \(f(x) = 6x + 11\) dollars for \(x\) miles, counting an \$11 pickup fee. Work backward: write the rule that turns a fare into miles, then apply it to a \$53 ride.
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Write \(y = 6x + 11\), then swap the letters: \(x = 6y + 11\).
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Solve for \(y\): \(x – 11 = 6y\), so \(f^{-1}(x) = \frac{x – 11}{6}\), the miles driven for a fare of \(x\) dollars.
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Evaluate at the fare: \(f^{-1}(53) = \frac{53 – 11}{6} = \frac{42}{6} = 7\). Check: \(f(7) = 42 + 11 = 53\).
The inverse is \(f^{-1}(x) = \frac{x – 11}{6}\), and a \$53 fare covers 7 miles.
Common mistake
Watch out
Nora claims the inverse of \(g(x) = 8x – 3\) is \(\frac{1}{8x – 3}\), reading the small \(-1\) as a reciprocal.
Why it is wrong
The symbol \(g^{-1}\) names the inverse, not \(\frac{1}{g}\). Test it: \(g(1) = 5\), yet \(\frac{1}{8(5) – 3} = \frac{1}{37}\) does not return to 1. Swapping and solving gives \(g^{-1}(x) = \frac{x + 3}{8}\), and \(g^{-1}(5) = 1\).
Related terms and lessons
- Algebra 1 · Building Functions and InversesLessonWorksheetQuiz
- Algebra 2 · Function Operations, Composition and InversesLessonWorksheetQuiz
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